I Riemann tensor in 3d Cartesian coordinates

[Nugatory]

Yea. The thing is that I would like to evaluate that integral without any dependence with another coordinate system. If I were to use polar coordinates on the integral, I would not even write down the metric in Cartesian coordinates.

If you want to take on that problem, you can...You'll find that integral in a table of integrals, and if you don't you could always use numerical methods to evaluate it. But why? Coordinate systems have no physical significance, so the only reason to use one that makes a problem unnecessarily difficult is because you enjoy the challenge. Physical insight doesn't come from brute-forcing your way through an awkward coordinate system, it comes from finding the coordinate system that makes a problem easy (Einstein's discovery of general relativity is the most spectacular example).

 

[Nugatory]

...because in this case we would have a straight line through the interior of the sphere, not the path through the surface of it.

But there's nothing wrong with that. You're working in a three-dimensional space to evaluate the length of a curve in that three-dimensional space so you're allowed to use points off the curve. The integration is a matter of drawing lines in that space that are not exactly on the curve but approximate it. That is an altogether different problem than evaluating the length of a curve on the surface of a two-dimensional sphere (although with proper choice of embedding you can make the two problems look very similar, and the answer comes out the same).

Throughout this thread I've been a bit unclear about what you're really looking for... I think you're looking for a way to work with the surface of a two-dimensional sphere when the points on the surface of the sphere are labelled with the same $x,y$ values that they have in a three-dimensional space in which the sphere has been embedded? If so, the following definition of coordinates on the surface of the sphere will do the trick:
$x=X$
$y=Y$
$z=\sqrt{R^2-X^2-Y^2}$
(I've used $X,Y$ for the coordinates on the surface of the sphere to avoid any ambiguity about when I'm referring to the mapping from points on the surface of the sphere onto $\mathbb{R}^2$ and when I'm referring to the mapping between points in the three-dimensional space onto $\mathbb{R}^3$).

Plug these into the formula for the induced metric, just as you would the latitude/longitude coordinates I used in the earlier example, and you will get (but note my earlier caution about wise people checking my algebra!) the metric:
$g_{XX}=1+\frac{X^2}{R^2-X^2-Y^2}$
$g_{YY}=1+\frac{Y^2}{R^2-X^2-Y^2}$
$g_{XY}=g_{YX}=0$
(Strictly speaking, these coordinates only apply to a hemisphere, not the full sphere. That's not surprising because we already know that a two-dimensional sphere cannot be covered by a single coordinate patch. The coordinate singularity along the curve $X^2+Y^2=R^2$ is an indication of the problem).

 

[davidge]

If you want to take on that problem, you can...You'll find that integral in a table of integrals, and if you don't you could always use numerical methods to evaluate it. But why? Coordinate systems have no physical significance, so the only reason to use one that makes a problem unnecessarily difficult is because you enjoy the challenge. Physical insight doesn't come from brute-forcing your way through an awkward coordinate system, it comes from finding the coordinate system that makes a problem easy (Einstein's discovery of general relativity is the most spectacular example).

It's just to know if it's possible to do that.

Throughout this thread I've been a bit unclear about what you're really looking for... I think you're looking for a way to work with the surface of a two-dimensional sphere when the points on the surface of the sphere are labelled with the same $x$,$y$ values that they have in a three-dimensional space in which the sphere has been embedded?

Yes. This motivated me to start the thread, although the title suggests I'm most interested on the Riemman tensor.

the following definition of coordinates on the surface of the sphere will do the trick:
$x=X\\ y=Y\\ z=\sqrt{R^2−X^2−Y^2}$

In this case we could not evaluate it through the path for which $z = 0$?

Just to know... Would the integral look like this?

$$ \int ds = \int g_{XX}dX^2 + \int g_{YY}dY^2 = \int \int dX dX + \int \int {\frac{X^2}{R^2 - X^2 - Y^2}}dX dX + \int \int dY dY + \int \int {\frac{Y^2}{R^2 - X^2 - Y^2}}dY dY $$

 

[Nugatory]

In this case we could not evaluate it through the path for which z = 0?

There is no $z$ in the two-dimensional surface of the sphere, so there is no such thing as a path for which z=0 and the question is meaningless. If you're working in the three-dimensional euclidean space using x,y, and z coordinates then there are curves along which z=0 and you can use the integral from your post #46 to find the length along such a curve. But note the way you wrote the bounds of the integration in that post - it's a single line integral, not a double integral across dx and dy so...

Just to know... Would the integral look like this?

No. You would have to do something analogous to what PeterDonis suggested in #45, writing the values of $X$ and $Y$ along the curve as functions of a single integration variable so you can evaluate the line integral along the curve.

 

[Nugatory]

although the title suggests I'm most interested on the Riemman tensor.

And now that you have the metric in these (remarkably difficult to work with) coordinates, you can calculate the components of the Riemann tensor in these coordinates.

 

[davidge]

There is no zzz in the two-dimensional surface of the sphere, so there is no such thing as a path for which z=0 and the question is meaningless. If you're working in the three-dimensional euclidean space using x,y, and z coordinates then there are curves along which z=0 and you can use the integral from your post #46 to find the length along such a curve

I used the points $(0\ , 5\ , 0)$ and $(1\ , \sqrt{24}\ , 0)$ and the relation $x^2 + y^2 = 25 = R^2$, $d\phi = dr = 0$, such that the line element in sph. coord. is $ds^2 = R^2 d\theta^2$.
$$ S = \int_{5}^{\sqrt{24}} dy \sqrt{1+ \frac{y^2}{R^2 - y^2}} = \int_{arcos(0.4)}^{\frac{\pi}{2}} R d \theta $$
The result I got was the same for the two integrals above.

And now that you have the metric in these (remarkably difficult to work with) coordinates, you can calculate the components of the Riemann tensor in these coordinates

I will work on it

 

[davidge]

I found two non-vanishing components of the Riemann tensor, namely
$$R^{y}\ _{xxy} = \frac{(2y^4/x^2)+3y^2}{(x^2+y^2)^2} = -R^{y}\ _{xyx}$$
Is it correct?

 

[PeterDonis]

Would the integral look like this?

No. What you wrote is an integral for $ds^2$, not $ds$. But that integral doesn't make sense, because you are integrating along a line element so you need to have just one integration variable, whereas you have two. In other words, you've written what's supposed to be the integral along a curve, but you're treating it as though it were an integral to determine an area.

The correct general form for the integral of the line element along a curve would be

$$
\int ds = \int \sqrt{ds^2} = \int \sqrt{g_{XX} dX^2 + g_{YY} dY^2}
$$

Which can't even be evaluated as it stands; you would need to find an equation for the curve, i.e., for $X$ in terms of $Y$ or $Y$ in terms of $X$ (or both of them in terms of some curve parameter), and use that equation to get the integrand into a form where it is a valid integrand with one integration variable.

 

[davidge]

@PeterDonis I constructed an integral following these steps and it worked. Please see my post #56.

 

[PeterDonis]

I constructed an integral following these steps and it worked. Please see my post #56.

Ah, ok, got it.

 

[PeterDonis]

using $\theta$ modulo $2\pi$ IS using multiple charts

I was hesitant in my response before but didn't have time to pin it down. Now I have; a while back we had a discussion about covering $S^1 \times R$ with a single chart, in which you posted:

The entire $S^1$ is an open interval, since it is the union of the interval $0 < \theta < \pi$ and $\frac{\pi}{2} < \theta < \frac{5 \pi}{2}$.

So it's a weird fact that $S^1$ cannot be covered by a single chart, but #S^1 \times R## can be.

This seems to suggest that we don't need to use multiple charts to have a coordinate, $\theta$ in this case, that is periodic.

 

[Nugatory]

I found two non-vanishing components of the Riemann tensor, namely
$$R^{y}\ _{xxy} = \frac{(2y^4/x^2)+3y^2}{(x^2+y^2)^2} = -R^{y}\ _{xyx}$$
Is it correct?

It's not promising that these expressions blow up at $X=Y=0$ where the coordinates and the manifold are well-behaved. Did you get these out of mathematica or equivalent, or do it by hand?

 

[Orodruin]

I was hesitant in my response before but didn't have time to pin it down. Now I have; a while back we had a discussion about covering $S^1 \times R$ with a single chart, in which you posted:
This seems to suggest that we don't need to use multiple charts to have a coordinate, $\theta$ in this case, that is periodic.

Strictly speaking, a chart must be a 1-to-1 mapping from the manifold to an open subset of $\mathbb R^n$ where $n$ is the dimension of the manifold. The typical one-chart atlas of the cylinder is to map it to a plane with one point removed - but you cannot use polar coordinates on that plane - you need to use the usual coordinates, or the map will not be 1-to-1.

 

[davidge]

It's not promising that these expressions blow up at $X = Y = 0$ where the coordinates and the manifold are well-behaved

Remember that I'm considering points along the surface of the sphere for which $z = 0$. So $x^2 + y^2 = R^2$, where $R$ is the radius of the sphere, and $X = Y = 0$ is not a possibility. The above expression for the two mentioned components would read, say in terms of y, $$\frac{2y^4 / (R^2 - y^2) + 3y^2}{R^4}$$

Did you get these out of mathematica or equivalent, or do it by hand?

I just calculated by hand the Cristoffel symbols and its derivatives and used it in the equation for the componets.

 

[Orodruin]

Remember that I'm considering points along the surface of the sphere for which $z = 0$. So $x^2 + y^2 = R^2$, where $R$ is the radius of the sphere, and $X = Y = 0$ is not a possibility. The above expression for the two mentioned components would read, say in terms of y, $$\frac{2y^4 / (R^2 - y^2) + 3y^2}{R^4}$$

I just calculated by hand the Cristoffel symbols and its derivatives and used it in the equation for the componets.

You are still doing it wrong then. You have to dismiss the idea of using the coordinates of the embedding space. On the part of the sphere where $z = 0$ in the embedding the coordinates $x$ and $y$ do not form a good coordinate system! You cannot use them as coordinates and compute stuff such as the components of the Riemann tensor, e.g., $R_{xyx}^y$.

 

[davidge]

the coordinates $x$ and $y$ do not form a good coordinate system

What do you mean by good coordinate system?

 

[Orodruin]

What do you mean by good coordinate system?

A continuous bijection from an open subset of $\mathbb R^2$ to a subset of the sphere.

 

[davidge]

A continuous bijection from an open subset of $\mathbb {R}^2$ to a subset of the sphere.

I don't see why it is not a bijection. Bijection means that each element on $\mathbb {R}^2$ is mapped on each point on the sphere, and no two points will be assigned to the same point on $\mathbb {R}^2$. Also, all points on the sphere will be mapped. Right? But the coordinates $x,y$ also do this job.

 

[stevendaryl]

I was hesitant in my response before but didn't have time to pin it down. Now I have; a while back we had a discussion about covering $S^1 \times R$ with a single chart, in which you posted:

This seems to suggest that we don't need to use multiple charts to have a coordinate, $\theta$ in this case, that is periodic.

Wow. I don't remember what I was thinking at that time. But you can come up with a mapping from the cylinder to the plane minus one point:

Map z,\theta to the point (x,y) with x = e^z cos(\theta), y = e^z sin(\theta). Every point on the cylinder is given a pair of coordinates, and distinct coordinates correspond to distinct points on they cylinder. But there is no point on the cylinder corresponding to the point x=y=0.

What I don't remember is whether a chart has to be simply-connected or not, or whether it's enough that it be an open set.

 

[Orodruin]

I don't see why it is not a bijection. Bijection means that each element on $\mathbb {R}^2$ is mapped on each point on the sphere, and no two points will be assigned to the same point on $\mathbb {R}^2$. Also, all points on the sphere will be mapped. Right? But the coordinates $x,y$ also do this job.

No they do not. At least not on the intersection of the embedding of the sphere in $\mathbb R^3$ and the plane $z = 0$ -- which is exactly where you are trying to make them work. If you pick a coordinate $x$ then the value of $y$ is completely determined. If it was a bijection you would be able to pick any values of $x$ and $y$ and they should correspond to a unique point on the sphere.

I will repeat one more time: Please stop thinking in terms of the embedding into $\mathbb R^3$.

 

[Orodruin]

What I don't remember is whether a chart has to be simply-connected or not, or whether it's enough that it be an open set.

The chart does not need to be simply connected. This is why you can pick the plane minus a point as a covering chart for the cylinder - as long as you do not use polar coordinates in the plane in which case you run into the same problem as you would in defining a covering chart for $S^1$.

 

[davidge]

No they do not. At least not on the intersection of the embedding of the sphere in $\mathbb {R}^3$ and the plane z=0##

If you pick a coordinate $x$ then the value of $y$ is completely determined. If it was a bijection you would be able to pick any values of $x$ and $y$ and they should correspond to a unique point on the sphere.

Can you explain me what actually makes the spherical coordinates be a bijection?

I will repeat one more time: Please stop thinking in terms of the embedding into $\mathbb{R}^3$.

Ok

 

[Orodruin]

Can you explain me what actually makes the spherical coordinates be a bijection?

Each combination of $\theta$ and $\varphi$ map to a unique point on the sphere. This is the definition of a bijection. Note that spherical coordinates are not a global chart - they only work locally away from the poles. This is fine as long as you stay away from the poles. If you use the coordinates x and y - you have to stay away from $z = 0$ where these coordinates are singular.

 

[davidge]

Each combination of $\theta$ and $\varphi$ map to a unique point on the sphere. This is the definition of a bijection

Ok, I got it.
Just another question: the Cristoffel symbols are:

$$\Gamma^{\mu}\ _{\nu \sigma} = \frac{1}{2}g^{\mu \rho}(g_{\nu \sigma, \rho} + g_{\rho \sigma, \nu} - g_{\nu \rho, \sigma}).$$
If the metric happens not to have an inverse $g^{\mu \rho}$, can we multiply and after divide both sides on the expression above by $g_{\mu \rho}$, getting

$$\Gamma^{\mu}\ _{\nu \sigma} = \frac{1}{2g_{\mu \rho}}(g_{\nu \sigma, \rho} + g_{\rho \sigma, \nu} - g_{\nu \rho, \sigma})?$$

 

[Orodruin]

Ok, I got it.
Just another question: the Cristoffel symbols are:

$$\Gamma^{\mu}\ _{\nu \sigma} = \frac{1}{2}g^{\mu \rho}(g_{\nu \sigma, \rho} + g_{\rho \sigma, \nu} - g_{\nu \rho, \sigma}).$$
If the metric happens not to have an inverse $g^{\mu \rho}$, can we multiply and after divide both sides on the expression above by $g_{\mu \rho}$, getting

$$\Gamma^{\mu}\ _{\nu \sigma} = \frac{1}{2g_{\mu \rho}}(g_{\nu \sigma, \rho} + g_{\rho \sigma, \nu} - g_{\nu \rho, \sigma})?$$

If the metric does not have an inverse it is not a metric.

 

[davidge]

If the metric does not have an inverse it is not a metric.

Then how to explain the fact that the line element $ds^2 = (1+ y^2/(R^2 - y^2))dy^2$ does correspond to $ds^2 = R^2 d\theta^2$ for the surface of a sphere, with $z = 0$? Or maybe I'm wrong about the definition of a metric?

 

[Orodruin]

Then how to explain the fact that the line element $ds^2 = (1+ y^2/(R^2 - y^2))dy^2$ does correspond to $ds^2 = R^2 d\theta^2$ for the surface of a sphere, with $z = 0$? Or maybe I'm wrong about the definition of a metric?

Because you are wrong and you are not using a good coordinate system as pointed out several times. You cannot expect to get something correct when you are not using an appropriate coordinate system.

 

[Nugatory]

Then how to explain the fact that the line element $ds^2 = (1+ y^2/(R^2 - y^2))dy^2$ does correspond to $ds^2 = R^2 d\theta^2$ for the surface of a sphere?

The calculations are different things acting on different mathematical objects: one is a calculation of the distance between two points in a two-dimensional space which has a particular metric tensor (note that the metric tensor, as opposed to its components, is independent of choice of coordinates!); the other is a calculation of the length along a curve in a three-dimensional space with a different metric tensor. It just so happens that you've embedded the two-dimensional space into the three-dimensional space in such a way that the two calculations will yield the same result.

 

[PeterDonis]

The typical one-chart atlas of the cylinder is to map it to a plane with one point removed - but you cannot use polar coordinates on that plane - you need to use the usual coordinates, or the map will not be 1-to-1.

Ah, got it.

 

[davidge]

You cannot expect to get something correct when you are not using an appropriate coordinate system.

What can we do to get the appropriate coordinate system without any knowledge about the shape of the space we are interested in? For example, suppose we did not know anything about the sphere. In such a case, is there a way of deriving the spherical coordinates?

To illustrate my question, suppose there are two-dimensional creatures living on the surface of the sphere. They see everything that exists (including themselves) liying on a plane, namely the surface of the sphere. Yet, (I think) they would detect curvature by the effects it causes, but they would be unable to use the spherical coordinate system to describe that curvature.

one is a calculation of the distance between two points in a two-dimensional space which has a particular metric tensor

the other is a calculation of the length along a curve in a three-dimensional space with a different metric tensor. It just so happens that you've embedded the two-dimensional space into the three-dimensional space in such a way that the two calculations will yield the same result.

I see

note that the metric tensor, as opposed to its components, is independent of choice of coordinates!

yea

 

[Orodruin]

To illustrate my question, suppose there are two-dimensional creatures living on the surface of the sphere. They see everything that exists (including themselves) liying on a plane, namely the surface of the sphere. Yet, (I think) they would detect curvature by the effects it causes, but they would be unable to use the spherical coordinate system to describe that curvature.

This is not true. We do use spherical cooordinates on the Earth's surface (longitude and latitude) and we can do so without reference to the embedding in three dimensional space. Your problem seems to be wanting to include the radius. The sphere is two dimensional and requires two coordinates only.

 

[davidge]

Your problem seems to be wanting to include the radius. The sphere is two dimensional and requires two coordinates only

The radius is included in the line element for the sphere $ds^2 = r^2(sin^2(\theta)d \varphi^2 + d\theta^2)$.

 

[Orodruin]

The radius is included in the line element for the sphere $ds^2 = r^2(sin^2(\theta)d \varphi^2 + d\theta^2)$.

That is a normalisation constant, not a coordinate.

 

[Nugatory]

o illustrate my question, suppose there are two-dimensional creatures living on the surface of the sphere. They see all that exists (including themselves) liying on a plane, namely the surface of the sphere. Yet, (I think) they would detect curvature by the effects it causes, but they would be unable to use the spherical coordinate system.

They would detect curvature by the effects that it causes: the interior angles of triangles would not add to 180 degrees; travel in a straight line in any direction eventually brings you back where you started; and other such effects. These would be enough to tell the mathematicians that the surface had the topology of a two-dimensional sphere, and that would be sufficient to justify the use of spherical coordinates.

(This discovery also vindicates the mathematicians, who had become tired of being incessantly asked what something as abstract as the topology of non-Euclidean manifolds was good for. Something similar happened with relativity - pseudo-Riemannian manifolds were just another interesting abstraction until experiment told us that spacetime is an example of one, and that branch of abstract mathematics suddenly became a foundation of real-world physics).

 

[Nugatory]

The radius is included in the line element for the sphere $ds^2 = r^2(sin^2(\theta)d \varphi^2 + d\theta^2)$.

You are confusing a normalization constant with a coordinate. The easiest way to see this is just to choose units of distance such that the radius of the sphere is equal to one - the constant $R$ disappears from both the calculations you do in the three-dimensional space and the two-dimensional space, but you still need the $r$ coordinate to use spherical coordinates in the three-dimensional space.

 

[davidge]

That is a normalisation constant, not a coordinate

Ah, ok. I realized this just before your reply.

(This discovery also vindicated the mathematicians, who had become tired of being incessantly asked what something as abstract as the topology of non-Euclidean manifolds was good for. Something similar happened with relativity - pseudo-Riemannian manifolds were just another interesting abstraction until experiment told us that spacetime is an example of one, and a branch of abstract mathematics suddenly became a foundation of real-world physics).

That is cool

They would detect curvature by the effects that it causes. These would be enough to tell the mathematicians that the surface had the topology of a two-dimensional sphere, and that would be sufficient to justify the use of spherical coordinates

So are you saying that there is no need for seeing e.g. a soccer ball to deduce its topology? In my example, the two-dimensional creatures would never completely seen a soccer ball, but they would still be capable of assiying angles $\theta, \varphi$ to it, although for him these angles could not be intuitively interpreted?

 

[Orodruin]

Ah, ok. I realized this just before your reply.That is coolSo are you saying that there is no need for seeing e.g. a soccer ball to deduce its topology? In my example, the two-dimensional creatures would never completely seen a soccer ball, but they would still be capable of assiying angles $\theta, \varphi$ to it, although for him these angles could not be intuitively interpreted?

Drawing two dimensional charts and comparing them and their overlaps, you can get a complete picture of how the Earth's surface behaves. On each chart, you only specify two coordinates.

 

[Nugatory]

So are you saying that there is no need for seeing e.g. a soccer ball to deduce its topology? In my example, the two-dimensional creatures would never completely seen a soccer ball, but they would still be capable of assiying angles $\theta, \varphi$ to it, although for him these angles could not be intuitively interpreted?

Well, we three-dimensional beings seem to have done a pretty good job of deducing interesting four-dimensional topologies, (including the conceptually more challenging pseudo-Riemannian solutions of general relativity where $ds^2$ between two different points can be zero or negative)... so I'd expect that our two-dimensional mathematicians would be able to figure out the soccer ball - especially if they lived on one.

You are also mistaken about the spherical coordinates not having an intuitive interpretation - they're just latitude and longitude. Indeed, even if our two-dimensional creatures had never birthed an abstract mathematician, they would discover these coordinates as soon as they started drawing straight lines on the surface of their planet or trying to describe the positions of things relative to one another. Pick two arbitrary points on the surface (on Earth we happened to choose the north pole and the Greenwich observatory). There is exactly one great circle through those points. Furthermore, for any any other point on the surface, there is exactly one great circle through that point and the arbitrarily chosen pole, and we can label that great circle by the angle it makes with the Greenwich observatory one where they meet at the pole. That's one coordinate, which we call "longitude". The distance to the point from the north pole along that great circle gives us the second coordinate, latitude... And all without ever messing with anything in the third dimension.

You should be able to convince yourself fairly quickly that if your only measuring instruments are strings stretched straight across the surface of the planet, and rulers to measure the length of these strings, you will be driven to discover latitude and longitude.

 

[davidge]

@Orodruin @Nugatory I think I got it.

we three-dimensional beings seem to have done a pretty good job of deducing interesting four-dimensional topologies, so I'd expect that our two-dimensional mathematicians would be able to figure out the soccer ball - especially if they lived on one.

You are also mistaken about the spherical coordinates not having an intuitive interpretation - they're just latitude and longitude

Oh yea, I see now

So to summarize the ideas of this thread:

(1) - To correctly get the curvature of a manifold, we must use a coordinate system that covers only the manifold in question; e.g. we need to use sph. coords. on a sphere, because all points it maps are on the sphere.

(2) - We can find a coord. system that satisfies (1), but maybe there's another coord. system more easy to work with. So we should always check if it's possible to define another coord. system even after we have found one that satisfies (1).

(3) - To learn about the curvature and any other intrinsic property of a manifold, we don't need to embed it in another higher-dimensional manifold. All that is to know can be found staying on it.

Am I getting these things correctly?

 

[pervect]

Yes, but some people require that a coordinate system must be a one-to-one map from an open set of R^N to an open set of the manifold. If (x,y) = (x,y+L), then it's not one-to-one. It's a picky point, but in the book that I read on differential geometry, there was that requirement.

With one chart that is an open set, you could cover everything on the cylinder except for one line. So you could take 0 < theta < 2pi, for instance, then you'd be missing the line theta=0. To cover that you'd need to introduce another chart to cover that missing line. This is I think perfectly possible, a manifold is defined by a collection of overlapping charts (sometimes called an atlas) meeting certain compatiblity conditions in the overlap region. It's a bit like the bound atlas of street maps where one cover say, a city's streets, with overlapping square maps. But it'd be a pain to do it all correctly, so I think physicists (as opposed to mathematicians) are more likely to take shortcuts.

 

[Orodruin]

With one chart that is an open set, you could cover everything on the cylinder except for one line.

As already stated several times in this thread, you can cover the entire cylinder with a single chart. Nothing requires the open set to be simply connected.

 

[pervect]

As already stated several times in this thread, you can cover the entire cylinder with a single chart. Nothing requires the open set to be simply connected.

I went back and reviewed my physics textbook (Wald) on that at point, and couldn't find anything definitive.

I suppose I'd need a textbook reference and/or (probably and) a lot more thinking to consider the case about a single chart overlapping itself, having never considered anything so strange before.

If I'm understanding you correctly, you're saying that we can apply the compatibility conditions that are required if any two charts O_a and O_b overlaping even if a=b, so that a chart is allowed to overlap itself as log as it satisfies the same compatibility conditions that we'd have for two different charts? At the moment, the idea is making my brain melt. I will say I am a lot more comfortable with having two different charts overlap than having a chart overlap itself.

 

[Ibix]

I went back and reviewed my physics textbook (Wald) on that at point, and couldn't find anything definitive.

Carroll's lecture notes imply it's possible to cover a cylinder with a single chart (last paragraph on p39, which is the 9th page of chapter 2: https://preposterousuniverse.com/wp-content/uploads/grnotes-two.pdf). He doesn't offer a solution, though (I think Orodruin posted the solution I'd seen before higher up this thread).

 

[PeterDonis]

(I think Orodruin posted the solution I'd seen before higher up this thread).

stevendaryl gave it in post #69.

 

[PeterDonis]

a chart is allowed to overlap itself

I don't think that's what Orodruin is saying. He's just saying that the open subset of $\mathbb{R}^N$ that is used in a coordinate chart does not need to be simply connected. For example, in the single-chart solution for the cylinder that stevendaryl posted in post #69, the open subset of $\mathbb{R}^2$ used in the chart is the plane minus a single point (i.e., all 2-tuples $(x, y)$ except $(0, 0)$), which is an open subset but is not simply connected. But the chart is still one-to-one; it doesn't overlap itself anywhere. (Note that you have to use the Cartesian coordinates $(x, y)$ on $\mathbb{R}^2$ for this to be true; it won't work if you use polar coordinates on $\mathbb{R}^2$, as Orodruin noted some posts ago.)

 

[Orodruin]

I would say something but Peter covered most of it in his post so let me just offer an explicit embedding of the cylinder in $\mathbb R^3$ using coordinates $(s,t)\neq (0,0)$:
$$
x = \frac{s}{r}, \quad y = \frac tr, \quad z = \ln(r)
$$
with $r = \sqrt{s^2+t^2}$. I think we can all agree that this is a continuous map from an open subset of $\mathbb R^2$ to the submanifold of $\mathbb R^3$ that we would typically refer to as a cylinder of radius one.

Just to add: The point is that the cylinder is homeomorphic to an open set in the plane while the circle is not homeomorphic to an open set in one dimension.

 

[stevendaryl]

I suppose I'd need a textbook reference and/or (probably and) a lot more thinking to consider the case about a single chart overlapping itself, having never considered anything so strange before.

If you look back over the posts, you will see that the one-chart coverage of the cylinder is just the Euclidean plane minus one point. There is no weirdness of a chart overlapping itself.

You map the point on the cylinder z, \theta to the point x = e^{z} cos(\theta), y = e^{z} sin(\theta).